Problem 31 - Algebra A Combined Function

Question

Solve. See Examples 1 through 5. $$ \frac{2 x}{x-2}+\frac{x}{x+3}=\frac{-5}{x+3} $$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{-9 \pm \sqrt{201}}{6}\).

Step 1: Identify the Least Common Denominator (LCD)

To solve the equation \( \frac{2x}{x-2} + \frac{x}{x+3} = \frac{-5}{x+3} \), we need a common denominator. The denominators are \((x-2)\) and \((x+3)\). Therefore, the least common denominator is \((x-2)(x+3)\).

Step 2: Multiply Through by the LCD

Multiply each term in the equation by the LCD \((x-2)(x+3)\) to eliminate the fractions:\[(x-2)(x+3) \left( \frac{2x}{x-2} \right) + (x-2)(x+3) \left( \frac{x}{x+3} \right) = (x-2)(x+3) \left( \frac{-5}{x+3} \right)\]This simplifies to:\[2x(x+3) + x(x-2) = -5(x-2)\]

Step 3: Distribute and Simplify

Distribute the terms in each product:- \(2x(x+3) = 2x^2 + 6x\)- \(x(x-2) = x^2 - 2x\)- \(-5(x-2) = -5x + 10\)Substituting back, we have:\[2x^2 + 6x + x^2 - 2x = -5x + 10\]

Step 4: Combine Like Terms

Combine like terms on the left side:\[2x^2 + x^2 + 6x - 2x = 3x^2 + 4x\]The equation becomes:\[3x^2 + 4x = -5x + 10\]

Step 5: Move All Terms to One Side

Add \(5x\) and subtract \(10\) from both sides to get all terms on one side:\[3x^2 + 4x + 5x - 10 = 0\]Simplify to:\[3x^2 + 9x - 10 = 0\]

Step 6: Solve the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 3\), \(b = 9\), \(c = -10\).Calculate the discriminant:\[b^2 - 4ac = 9^2 - 4(3)(-10) = 81 + 120 = 201\]The roots are:\[x = \frac{-9 \pm \sqrt{201}}{6}\]

Step 7: Check the Solutions

Check if the solutions satisfy the original equation and don't make the denominator zero. Both solutions should be checked to ensure that neither \(x = 2\) nor \(x = -3\), which would make the original denominators zero. Solve \(x = \frac{-9 + \sqrt{201}}{6}\) and \(x = \frac{-9 - \sqrt{201}}{6}\) and ensure neither value is \(2\) or \(-3\). Neither solution simplifies to these values, so both are valid.

Key Concepts

Least Common Denominator

The first step in solving rational equations is finding the Least Common Denominator (LCD). This is crucial because it allows us to eliminate fractions by converting them into polynomials. Consider the equation: \( \frac{2x}{x-2} + \frac{x}{x+3} = \frac{-5}{x+3} \). The denominators here are \((x-2)\) and \((x+3)\).
To find the LCD, identify all unique factors in the denominators. In this equation, these factors are \((x-2)\) and \((x+3)\). The LCD is, therefore, the product of these factors, or \((x-2)(x+3)\).
Having established the LCD, multiply each term by this expression to clear the fractions and proceed to solve the resulting polynomial equation.

Quadratic Formula

The quadratic formula is a powerful tool to solve quadratic equations of the form \( ax^2 + bx + c = 0 \). It is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula finds all possible values of \(x\) that satisfy the equation. In our solution, after eliminating fractions, we derived the quadratic equation \(3x^2 + 9x - 10 = 0\).
Here, \(a = 3\), \(b = 9\), and \(c = -10\). Substituting these into the formula will give us the two solutions of the equation. It's essential for students to remember the formula and know how to substitute the values correctly to avoid mistakes.

Discriminant

The discriminant is a part of the quadratic formula found under the square root sign: \(b^2 - 4ac\). It tells us about the nature of the roots of the equation. Here's why it's useful:

• If the discriminant is positive, the quadratic equation has two distinct real roots.
• If it's zero, there is one real root (a repeated root).
• If negative, the roots are complex and not real.

In our equation, the discriminant is calculated as: \( 9^2 - 4(3)(-10) = 201 \).
Since 201 is positive, there are two distinct real roots. Understanding the discriminant helps in anticipating the types of solutions and is important for interpreting the results.

Simplifying Expressions

Simplifying expressions involves reducing equations or expressions to their simplest form while keeping their value unchanged. This is key when solving rational equations where terms can become quite complex. In our example, after multiplying through the LCD, we obtained:
\(2x(x+3) + x(x-2) = -5(x-2)\).
Distributing terms gives us:
- \(2x(x+3) = 2x^2 + 6x\)
- \(x(x-2) = x^2 - 2x\)
- \(-5(x-2) = -5x + 10\)
Collect like terms and simplify again to: \(3x^2 + 9x - 10 = 0\).
This simplification shows how initial complex expressions can be made more manageable, helping to solve the equation efficiently. Always check that simplifications maintain accuracy and follow these processes step by step to enhance understanding.